Question

Find the value of x +9,x-6,4 are the three terms of a geometrical progression and calculate the fourth term of progression in each case

Answer 1

Step-by-step explanation:

$t1 = x + 9 \: \: \: \: t2 = x - 6 \: \: t3 = 4 \\ in \: g.p. \\ \\ \frac{t2}{t1} = \frac{t3}{t2} \\ \frac{(x - 6)}{(x + 9)} = \frac{4}{(x - 6)} \\ {(x - 6)}^{2} = 4(x + 9) \\ {x }^{2} - 12x + 36 = 4x + 36 \\ {x}^{2} - 12x - 4x = 36 - 36 \\ {x}^{2} - 16x = 0 \\ {x}^{2} = 16x \\ devide \: by \: x \: both \: sides \: \\ \frac{ {x}^{2} }{x} = \frac{16x}{x} \\ x = 16$

geometric progression is 25,10,4

Tn=ar^(n-1)

a=25, r=0.4=4\10

$t4 = 25 \times ( { \frac{4}{10}) }^{(4 - 1)} \\ = 25 \times { (\frac{4}{10}) }^{3} \\ = 25 \times \frac{64}{1000} \\ = \frac{1600}{1000} \\ t4 = 1.6$